To and Fro（字符串水题）

To and Fro

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Problem Description

Mo and Larry have devised a way of encrypting messages. They first decide secretly on the number of columns and write the message (letters only) down the columns, padding with extra random letters so as to make a rectangular array of letters. For example, if the message is “There’s no place like home on a snowy night” and there are five columns, Mo would write down

t o i o y
h p k n n
e l e a i
r a h s g
e c o n h
s e
m o t
n l e w x

Note that Mo includes only letters and writes them
all in lower case. In this example, Mo used the character ‘x’ to pad the message
out to make a rectangle, although he could have used any letter.

Mo then
sends the message to Larry by writing the letters in each row, alternating
left-to-right and right-to-left. So, the above would be encrypted
as

toioynnkpheleaigshareconhtomesnlewx

Your job is to recover for
Larry the original message (along with any extra padding letters) from the
encrypted one.

 

Input

There will be multiple input sets. Input for each set
will consist of two lines. The first line will contain an integer in the range
2. . . 20 indicating the number of columns used. The next line is a string of up
to 200 lower case letters. The last input set is followed by a line containing a
single 0, indicating end of input.

 

Output

Each input set should generate one line of output,
giving the original plaintext message, with no spaces.

 

Sample Input

5

toioynnkpheleaigshareconhtomesnlewx

3

ttyohhieneesiaabss

0

 

Sample Output

theresnoplacelikehomeonasnowynightx

thisistheeasyoneab

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 int main()
 {
     int i,n;
     char ch;
     char a[][];
     char b[];
     bool flag=;
     int p=,j=;
     while(cin>>n&&n)
     {
         int k=;
         memset(b,,sizeof(b));
         memset(a,,sizeof(a));
         getchar();
         gets(b);
         int len=strlen(b);
         while(k<=len)
         {
             if(!flag)
             {
                 for(i=;i<n;i++)
                     if(k<=len)
                         a[p][i]=b[k++];
                 flag=;
                 p++;
                 continue;

             }
             if(flag)
             {
                 for(i=n-;i>=;i--)
                     if(k<=len)
                         a[p][i]=b[k++];
                 flag=;
                 p++;
                 continue;
             }
         }
         for(i=;i<n;i++)
         {
             for(j=;j<p-;j++)
             {
                 printf("%c",a[j][i]);
             }
         }
         cout<<endl;
         p=;
     }
 }

 #include <stdio.h>
 #include <string.h>
 int main()
 {
     char c,s[][];
     int i,j,k,n,f,count;
     while(scanf("%d",&n),n){
         f=count=;
         for(k=;f==;k++)
             for(i=;i<=n&&f==;i++){
                 scanf("%c",&c);
                 if(c=='\n') f=;
                 else{
                     count++;
                     if(k%==) s[k][i]=c;
                     else s[k][n-i+]=c;
                 }
             }
         for(i=;i<=n;i++)
             for(j=;j<=count/n;j++)
                 putchar(s[j][i]);
         putchar('\n');
     }
     return ;
 }